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I was trying to compute the value of $\sum_{n=0}^\infty n(n-1)x^n$ and realizing this is the second derivative of $\sum_{n=0}^\infty x^n = \frac{1}{1-x}$ (whenever $|x|<1$) I proceed to take derivatives. The first derivative of the power series is $$\sum nx^{n-1}=\frac{1}{(1-x)^2}$$

And this is when I get in trouble. Rewriting this last term as $(1-x)^{-2}$, I compute its derivative as $(-2)(1-x)^{-3}(-1)=\frac{2}{(1-x)^3}$, on the other hand $\frac{1}{(1-x)^2} = \frac{1}{(x-1)^2}$ so it seems that the derivative of $(x-1)^{-2}$ is $(-2)(x-1)^{-3}(1)=\frac{-2}{(x-1)^3}$.

While we're at it.... I feel like this "paradox" is probably related to the fact that while $\cos(x)=\cos(-x)$ it is not the case that $(\cos(-x))'=\sin(x)$. I'm not sure how to make sense of this last fact either, but I suspect direct application of the derivative's definition will resolve this .... it is still unsettling.

How can we resolve these paradoxes?

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    $\frac{-2}{(x-1)^3}=\frac{2}{-(x-1)^3}=\frac{2}{(-1)^3(x-1)^3}=\frac{2}{(1-x)^3}$ – 2017-02-19
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    Okay. But why is the derivative of $\cos$ not $\sin$? – 2017-02-19
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    And $(\cos(-x))'=(\cos(x))'$ for the same reason, so all is well. – 2017-02-19
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    The derivative of $\cos(x)$ was never $\sin(x)$ in the first place. It's $-\sin(x)$. – 2017-02-19
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    B.t.w. every should know the derivative of $ \dfrac1{x^n}=-\dfrac n{x^{n+1}} $ without having to convert to a negative exponent. – 2017-02-19

2 Answers 2

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There are some paradoxes in mathematics, but not in your examples. $$-\frac{2}{(1-x^3)} = \frac{2}{(x^3-1)}$$ And $$(\cos(x))'=-\sin(x)$$ while $$(\cos(-x))'=-\sin(-x)\cdot(-x)'= \sin(-x)=-\sin(x)$$

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Consider \begin{align} f(x)&=\frac{1}{(1-x)^2}=(1-x)^{-2} \tag{1}\\ &=\frac{1}{(x-1)^2}=(x-1)^{-2} \tag{2} \end{align} Derivative, first form: $$ f'(x)=-2(1-x)^{-3}(-1)=\frac{2}{(1-x)^3} $$ Second form: $$ f'(x)=-2(x-1)^{-3}(1)=-\frac{2}{(x-1)^3} $$ Since $(1-x)^3=-(x-1)^3$, there's no contradiction nor paradox.